Question

# Solve the initial value problem displaystyle{leftlbracebegin{matrix}{y}text{}+{16}{y}= cos{{left({4}{t}right)}}{y}{left({0}right)}={1}{y}'{left({0}right)}={1}end{matrix}right.} using the Laplace transform.

Laplace transform
Solve the initial value problem $$\displaystyle{\left\lbrace\begin{matrix}{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}}\\{y}{\left({0}\right)}={1}\\{y}'{\left({0}\right)}={1}\end{matrix}\right.}$$ using the Laplace transform.

2020-10-24
Step 1
Given: $$\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Step 2
Explanation:
$$\displaystyle{y}\text{}+{16}{y}= \cos{{\left({4}{t}\right)}},{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={1}$$
Taking Laplace transform on both sides
$$\displaystyle{L}{\left[{y}{''}+{16}{y}\right]}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}$$
$$\displaystyle{L}{\left({y}\text{}{\left({t}\right)}\right)}+{L}{\left({16}{y}{\left({t}\right)}\right)}={L}{\left[ \cos{{\left({4}{t}\right)}}\right]}\ldots{\left({i}\right)}$$
We know that if $$L(y(t))=y(s)$$
then $$\displaystyle{L}{\left({y}{''}{\left({t}\right)}\right)}={s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}$$
$$\displaystyle{L}{\left[ \cos{{\left({a}{t}\right)}}\right]}=\frac{s}{{{s}^{2}+{a}^{2}}}$$
Using these in equation (i), we get
$$\displaystyle{s}^{2}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{16}{y}{\left({s}\right)}=\frac{s}{{{s}^{2}+{4}^{2}}}$$
Using the given initial conditions $$y(0)=1, y'(0)=1$$ then
$$\displaystyle{\left({s}^{2}+{16}\right)}{y}{\left({s}\right)}-{s}{\left({1}\right)}-{1}=\frac{s}{{{s}^{2}+{4}^{2}}}$$
$$\displaystyle{\left({s}^{2}+{4}^{2}\right)}{y}{\left({s}\right)}={1}+{s}+\frac{s}{{{s}^{2}+{4}^{2}}}$$
$$\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}{\left({s}^{2}+{4}^{2}\right)}}}$$
$$\displaystyle{y}{\left({s}\right)}=\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{\left({s}^{2}+{4}^{2}\right)}^{2}}}$$
Taking inverse Laplace transform on both sides
$$\displaystyle{L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}+\frac{s}{{{s}^{2}+{4}^{2}}}+\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}$$
$$\displaystyle{y}{\left({t}\right)}={L}^{ -{{1}}}{\left(\frac{1}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{{s}^{2}+{4}^{2}}}\right)}+{L}^{ -{{1}}}{\left(\frac{s}{{\left({s}^{2}+{4}^{2}\right)}^{2}}\right)}$$
$$\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{{2}\cdot{4}}}{t} \sin{{\left({4}{t}\right)}}$$
$$\displaystyle=\frac{1}{{4}} \sin{{\left({4}{t}\right)}}+ \cos{{\left({4}{t}\right)}}+\frac{1}{{8}}{t} \sin{{\left({4}{t}\right)}}$$